Monday, 24 April 2017

What Is Quadratic Equation And How To Solve Them for IBPS PO Exam

quadratic equation

As discussed previously, we understand how important a role the topic, ‘equations’ plays in solving most of the questions in the Quantitative Aptitude section of any major competitive exam, such as SBI PO bank exam, IBPS PO bank exam, SSC CGL 2017, and more. We have also learnt the different types of equations, what linear equation is and how to solve them. 

Here, let us talk about what is quadratic equation and how to solve them easily.

What is Quadratic Equation?

In a Quadratic Equation, all the powers of the variable are positive and the highest power is 2. Hence, we can also say that it is a second-degree polynomial which is equated to zero. 
QUADRATIC EQUATIONS
Explanation: Since the highest power of x in a Quadratic Equation is 2, there will always be 2 values of the variable that satisfy the equation, unlike a linear equation, where x can have only one unique value.

These 2 values of the variable x are called the roots of the equation.

How To Find The Roots Of A Quadratic Equation?

There are 2 methods to find out the two roots of a quadratic equation. 

First Method: Standard Method
QUADRATIC EQUATIONS
Second Method: Factorization
QUADRATIC EQUATIONS
In factorization method, a quadratic equation is written in terms of two simple linear equations. Sometimes, it may not be simple to solve a quadratic equation in this method. In which case, we can always solve the equation in the standard method.

Example 1:
Quadratic equation with 2 variable: 
x2 - 5x + 6 = 0
Solution:
x2 - 5x + 6 = 0
Here, a = 1, b = -5, c = 6

First Method: Standard Formula
Second Method: Factorization
Example 2:
Solve the given equation and establish the relationship between x and y.
x2 – 10x +21 = 0
y2 – 16y + 63 = 0
Solution:
We have to solve these equations separately and find out the values of x and y.

First Equation:
x– 10x +21 = 0

Since we know that the standard method is quite straightforward, let us try and find the variables x and y using factorization method.

Second Equation:
y– 16y + 63 = 0
Comparision of variables:
To establish the relationship between x and y, we must compare both the values of x with both the values of y. 

Comparing x and y as required, we find that:
Since comparing the 2 values of x with the 2 values of y, we can establish that:
In some cases, we might find that 2 of the comparisons might say that x>y, while the other 2 say that x<y. In such cases, it can be said that the relationship between x and y can not be established. 

Example 3:
Solve the given equation and establish the relationship between x and y.
17x+ 48x = 9
13y2 = 32y - 12

Solution:
We have to solve these equations separately and find out the values of x and y.

First Equation:
17x+ 48x = 9
17x+ 48x - 9 = 0
Second Equation:
13y= 32y - 12
13y2 - 32y + 12 = 0
Comparision of variables:
To establish the relationship between x and y, we must compare both the values of x with both the values of y. 
Since the value of x is negative,
Since the value of x is less than 1 (as in case of any fraction),
Now to compare the two remaining fractions:

Since comparing the 2 values of x with the 2 values of y, we can establish that:

Point to remember: 
How to easily split the constant 'b': 

1. We can use the trail and error method.
2. Find b1 such that bis a multiple of a AND b2 such that b2 is a factor of c.


In the first equation of Example 3, to split 48 (b), we can simply start checking with the multiples of 17 (a). 
48 = 34 + 14 but 34 * 14 ≠ 153
48 = 51 - 3 and 51 * 3 = 153
Where 34 and 51 are multiples of 17

In the second equation of Example 3, to split 32 (b), we can check the multiples of 13 (a).
32 = 26 + 6 and 26 * 6 = 156

This simplifies solving quadratic equation in the factorization method. However, to be able to perfect this method, you need to practise a great deal.


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